CE 311K Introduction to Computer Methods Daene McKinney Introduction .
Introduction CE 408_Environ.ii
Transcript of Introduction CE 408_Environ.ii
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Sanitary Engineering OR Sanitation
INTRODUCTION
Sanitary Engineering also known as Sanitation Engineering is the branch of Environmental
Engineering in which the basic principles of science and Engineering are applied to the problem of
water pollution control. The main purpose of sanitary engineering is to maintain such an
environment that not affect the public health in general. Thus sanitary Engineering aims at the
creation of such conditions of living which will not result into serious outbreak of epidemic. It is thus
a preventive measure for the preservation of health of the community in general and individual in
particular.
Sanitary Engineering is the start point of the end of water supply. It is thus the operation of collecting
wastewater from different generating sources, proper treatment before its disposal to water bodies
or reuse for different purposes. If untreated wastewater is allowed to accumulate, the decomposition
of the organic materials it contains can lead to the production of large quantities of gases causing
unfavorable smell in atmosphere. In addition wastewater usually contains numerous pathogenic
microorganisms came from human being, animals or from industrial wastes. Similarly, the presence
of nutrients in waste water can stimulate the growth of aquatic plants, which contain toxic materials.
Thus the removal of wastewater from its source of generation, followed by treatment and disposal is
not only desirable but also essential for a civilized society.
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BASIC DEFINITIONS
SEWAGE OR WASTEWATER: The liquid waste from residential buildings, offices, other buildings and
institutions etc. as well as industrial wastewater and storm sewage including infiltration and inflow.
SEWER: The underground conduits or drains through which sewage is conveyed. Sewer are generally
closed, but not flowing full. The full flowing sewer are known as force main as the flow is under pressure.
DOMESTIC OR SANITARY SEWER: The sewage which originates of dwellings, commercials or
industrial facilities, and institutions. The sewer known as sanitary sewer.
INDUSTRIALS WASTES: The liquid discharges from industrials processes like manufacturing and food
processing. The sewer is industrial sewer.
STORM SEWAGE: Storm sewage is the flow derived from rainfall events and introduced into sewers
intended for its conveyance. The sewer used for this purpose is known as storm sewer.
INFILTRATION: Infiltration is the water enter into sewer from the ground through leaks.
INFLOW: It is the water enter to the sewer from surface source through cracks in manholes and open
cleanouts. Inflow usually occurs during runoff.
EXFILTRATION: The leakage of water out of sewer to ground is exfiltration.
GARBAGE: It is used for dry refuse of town containing organic, inorganic solids, combustible, non
combustible, putrescible and non putrescible substances. It includes sweeping from homes, streets,
markets, offices, gardens and other public places. Waste papers, leaves, grasses, decaying fruits and
vegetable etc. with small quantities of sand, silt, clay, debris are all garbage.
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Basic DefinitionsIt is collected separately from sewage and sullage and disposed off separately
SULLAGE: It is the wastewater resulting from personal washing, bathing, laundry, food preparation
and cleaning of utensils.
. SEWERAGE: The collection, treatment and disposal of liquid waste.
SEWERAGE WORK OR SEWAGE WORK: It include all appurtenances required for the collection,
treatment and disposal. collection work are provided for collecting sewage from different points of
occurrences and conveying sewage to any desired points with the help of a sewer system.
Houses connection, laterals, branch sewer, sub main, main, trunk
Treatment and Disposal
INVERT: The lowest level or surface of sewer.
COMBINED SEWAGE :A combination of sanitary and storm sewage with or without industrial waste.
CRUDE OR RAW SEWAGE: not treated
DILUTE OR WEAK SEWAGE: Sewage containing less suspended solids.
FRESH SEWAGE: fresh produced sewage.
DRY WEATHER FLOW: Normal flow of sewage during dry season. When there is no rainfall for at
least three days continuously, then the flow is Dry Weather Flow (DWF).
Wet Weather Flow: The flow that occurs during a rainy season is Wet weather flow. I include
domestic and runoff (WWF).
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TYPES OF SEWERS w.r.t WASTEWATER
Sanitary Sewer:
Sewer which carried sanitary sewage i.e. W.W originating from a municipality including DOMESTIC and
INDUSTRIAL wastewater.
Storm Sewer:
It carries storm sewage including surface run off and street washes.
Combined Sewer:
It carries domestic, industrial and storm sewage.
House Sewer:
A pipe conveying wastewater from an individual structure to a common sewer or point of disposal
Lateral Sewer: It receive discharge from house sewers.
Sub main sewer: It receive discharge from one or more laterals
Main / Trunk Sewer: Receive discharge from two or more sub mains.
Outfall Sewer
Receive discharge from all collecting system and convey it to the point of final disposal (e.g a
water body etc.)
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Basic DefinitionsCLASSIFICATION OF SEWERAGE SYSTEM
A: Separate: When sanitary sewage and storm wastewater flow in separate sewerage system. Then it is
known as separate system and the sewers are terms as separate sewers.
ADVANTAGES: less load on treatment unit ; natural water or storm water not polluted; small size of sewer
required ; economical; discharged into natural water bodies without treatment.
DISADVANTAGES: cleaning and maintenance difficult ; two sets of sewers required ; storm sewer dry
normally in dry season and deposition of sediments occurs.
B: Combined : When all the waste water from sanitary sewage, industrial sewage and storm water is
removed in one set of sewer. Such sewerage system is combine sewerage system and the pipe to carry
such sewage is combine sewer.
Advantages : easy to clean; maintenance cost is less; one set of sewer and is economical
Disadvantages: silted; load increases on treatment plant: larges size of sewer; uneconomical if pumping
required; overflow occasionally
C: Partially combined or Partially separate system: the sewage and storm water of buildings are in one
set and storm water from roads, streets, pavement are carried out by an other sewer.
Force Main: A pressurized sewer lines which convey sewage from a pumping station to another main or
treatment unit.
Relief sewer a sewer which has is built to carry a portion of the flow in a system to other sewer
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Quantity of sewage
Quantity of sewage: Dry weather flow and storm water
Dry weather flow: Domestic and industrial waste (Base flow)
Dry weather : Infiltration and exfiltration can be prevented up to certain level by water tight joints of sewer
Infiltration depends on
(a) subsoil water level (b) length of sewer (c) nature& type of soil through which sewer laid (d) size of sewer .
Infiltration can be expressed as (i) liters/day-hect. Of area (ii) liters-day-km length of sewer (iii) liters/day-cm
dia – km length of sewer
• Nature of industries
Population: 70 to 80 % of water supply; losses due to consumption; evaporation in industries; lawn
irrigation etc.
Two controlling factors in design of sewer: Maximum and Minimum rate of sewage. Maximum for design
purpose & minimum for controlling of sediment:
• The variation of maximum, average and minimum is sometime great and need a particular multiplying factor
known as Peak factor and is defined as = Peak/average flow rate; for residential area
• P.F or M= 1 + 14/(4 +P) (in thousands); According to WASA (water and sanitary Agency) M= 2.3 X average
flow. In normal conditions M= 4 for laterals; 2.5 for main and trunk and 2.0 for combined sewer
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Example
• The residential area of a city has a population density of 15000 persons / Km2 and
an area of 120000 m2. If average water consumption is 400 lpcd. Find the
average and maximum sewage flow in. m3 / day
Given information:
Pop. Density = 15000 per / km2 Area = 120000 m2
Avg. Water Consumption= 400 lpcd
Required: Average & maximum sewage flow
Solution:
Total population = 15000 x 120000 / (1000)2 = 1800 persons
Avg. Sewage flow = 1800 x 400 x 80/100 = 576000 l/day = 576 m3/day
Peak factor: M = 1 + 14/(4 +P) (in thousands) M=1 + 14/(4 +1.80) = 3.62
Max sewage flow = 3.62 x 576 = 2085 m3/day =
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Sewage Quantity
• Storm Water Sewage Estimate:
• Rational Formula
• Rational formula: Simplest formula used for area ≥0.5 km2≤3.0 km2 catchment's area. The general form of
Rational formula is Q = 0.0028 C i A
where Q = peak discharge in cubic meter per second; C = Runoff coefficient based on surface storage,
infiltration and evaporation less for pervious and more for impervious terrain ; i=rainfall intensity in mm hr -1;
A = catchment area in hectares
• For impervious surfaces C= 0.175t 1/3 or C= t /(8+t)
• For pervious surfaces C= 0.3xt/(20+t) where t is the duration of storm in minutes
Time of concentration method
• Time of concentration consists of two parts
(i) inlet time or overland flow time or time of equilibrium (Ti): the time taken by water to flow overland from the
critical point up to the point where it enters the drain. Mathematically Ti =( 0.885 L3 /H)0.385 where Ti = Inlet
time in hours; L = length of overflow in km from critical point to drain mouth and H = total fall of level fromcritical point to the mouth of drain in meters
(ii) The channel flow time (Tf ): The time taken by the water to flow in the drain channel from the mouth to the
considered point. Tf =Length of drain/velocity of drain
• Time of concentration or time of equilibrium= Ti+Tf Izzard equation te=526.76kL(1/3)ie
(-2/3)
where k =2.76x10-5xie+c /s1/3 where L= distance of flow (m) S = slope; c= retardance coefficient ; ie=excess
rainfall; mmh-1
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Example
A watershed to contribute an urban storm drainage is 50 hectares, of which 30 ha. having C value of 0.36
with 7% slope, and 20 ha. of 0.5 with 3% slope. The length of run is 720 meters. Calculate peak runoff if
rainfall intensity is 15 cm/hour. Velocity of flow is 1.2 m/s
Given Information: Area = 50 hectares of which 30 ha. Having C = 0.36 and 20 ha = 0.5 I= 15 cm/hr.
Required : Peak Discharge
Solution: The weighted value of C = (30* 0.36/50) + (20*0.5/50) = 0.415 = 0.416
Time of concentration = Length of run/ velocity = 720/(1.2 * 60 ) = 10 minutes
Rational formula for peak discharge Q = 0.0278 C I A = 0.0278 * 0.416 * 15 *50 = 8.65 m3/sec Answer
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Storm water sewage
SOIL CONSERVAT ION SERVICES (SCS)TECHNIQUE:
Soil conservation techniques is based on Curve Number (CN), which is a runoff coefficient and depends on
soil type, antecedent moisture. Hydrological soil group are classified as A,B,C &D.
Group A Lowest runoff potential . It consists of deep, well to excessively drained sands or gravels
with very little clay and silt, also deep, rapidly permeable soil. The infiltration rate is very high, and
minimum or no runoff occur from this hydrological group.
Group B Moderately low runoff potential . Such group is mostly sandy soil with less deeper than
group A. It also less aggregated than A. It consists of moderately deep; moderately well to drained coarse
soil. This group of soil have less average infiltration and more runoff than Group A.
Group C Moderately high runoff Potential : It consists of shallow soil and soil containing
considerable clay and colloids. This group has below average infiltration than group D but more than B.
The runoff concentration for such soil is more as compared to B but less than D.
Group D Highest runoff Potential : This group include mostly clay of high swelling properties. This
cause the infiltration rate minimum and high runoff .This group also include some shallow soil with nearly
impermeable sub horizon near the surface causing to minimize the infiltration and resulting of more
runoff.
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Example
Example: Calculate the size and slope of a sanitary trunk sewer serving a population of 0.4
million. Water consumption is estimated to be 300 litre per capita day. Pipe used is RCC. Take
infiltration as 10% of average sewage flow. Assume average velocity as 0.6 m/sec and n= 0.013
Given Data: Population=0.4 million; water consumption= 300lpcd; Infiltration =10%
Required: Size & slope of sewer.
Solution: Water Supply = Population x per capita demand
= 0.4 x 1,000,000 x 300 = 120000 m3 /day= 1.388 m3 sec -1.
Dry weather flow: Average Sewer flow = 1.388 x 80/100 = 1.11 m3 / sec= 96000 m3 / day
P. F = 1+ ( 14/(4 +P)) = 1+ ( 14 /(4+ (400000/1000))) = 1.58 when population is not given thencalculation is as
Peak Flow = 2.3 x 96000 = 220800 m3 /day; (in case of Population = 1.58 *96000 = 151680)Infiltration = 10/100 x 96000 = 9600 m3 /day (15168 m3 /day )Design Flow = Peak Flow + infiltration
= 220800 + 9600 = 230400 m3
/ day= 2.67 m3
/ sec (166848 m3
/ day = (1.93 m3
/s)Dia of sewer:
Q = AV = 230400 x 1 / 24 x 60 x 60 = A x 0.6 (1.93 = A * 0.6) => ( A = 3.22)
A = 4.45 m2
Dia = 2.38 m (Dia = 2.02 mSlope of sewer:
V = 1/n R 2/3 S1/2
0.6 = 1/0.013 x (2.38/4) 2/3 (S)1/2 =>S = 1.2157 x 10-4 ( 1. 51 x 10-4
S = 0.0001215 mm-1
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ProblemsProblem #13.1: A residential urban area has the following proportions of different land use; roofs 25%; asphalt
pavement 14%; concrete sidewalk, 5%; gravel driveways, 7% grassy lawns with average soil and little slope,
49%. Compute an average runoff coefficient using the values in Table 13-2.
• Solution
• Types of surface %age C value Product
roofs 25% 0.70-0.95 0.175-0.2375
Asphalt pavement 14% 0.85-0.90 0.119-0.1260
Concrete sidewalk 5% 0.85-0.90 0.0425-0.045
Gravel driveways 7% 0.05-0.10 0.0035-0.007
Grassy lawns 49% 0.13-0.17 0.0637-0.0833
--------------------------------------------------------------------
Total 100 0.4037-0.4988
Average C = 0.45125
Problem#13.2: An urban area of 100,000 m2 has a runoff coefficient of 0.45. Using a time concentration of 25
minutes and data of Fig. 13-1, compute the peak discharge resulting from a 10-year storm.
Solution:- Area =A 100,000 m2 = 10 hectares; Tc= 25 minutes (from fig.13.1 at 25 minutes i= 120 mm h-1
C= 0.45
Therefore According to Rational Formula Q = 0.0028*C*i*A = 0.0028* 0.45 * 120 *10 =1.512 m3 sec -1
Answer
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Problem#13.5• A watershed of 4,000,000 m2 has a present CN of 70 and an average slope of 3%. Development
will modify 70% of the hydraulic length, increase the impervious area to 40%, and increase CN to
80. compute the present and future peak discharge from a 100mm 24-hr.storm.
• Given information:- Area =4,000,000 m2 CN= 70 Slope = 3% Modification causedarea = 70% CN = 80 Rainfall = 100mm-24hr storm
• Required: Present and future Peak discharge
• Solution: Peak discharge with CN 70 and 100 mm = 33.3 mm/24hr
So Peak discharge = 4,000000 * 33.3 mm*m/1000mm * 1/24hr *hr/60mint*mint/60 secs =1.542 m3 /sec• Development % CN Product
70 80 560030 70 2100
--------------------------------------------------------- Average CN = 7700/100 = 77By Interpolation (X-X1 / X2-X1) =( Y-Y 1 / Y 2-Y 1) => (77-75) / (80-75)= (Y-41.8/(51.0-41.8)
= 2/5 = Y- 41.8/9.2 = 9.2*2 = 5Y -209 =Y => 18.4+209/5 = 227.4/5 = Y => 45.48 mm
Total Peak discharge = 4,000000 m2 * (45.48 mm/1000 mm/m) * (1/24hr*hr/60 minutes*minutes/60 secs)
= 2.106 m3 sec -1 Answer
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Problem
A 225 mm sewer is to flow full at velocity of 0.75 m/s
– A: What is its maximum capacity and what minimum grade it should be laid if coefficient ofroughness is 0.013. How many people it can serve if the Peak average per capita flow is 300lpcd. Ignore infiltration
– B: How many additional people the sewer can cater if slope is doubled
Given Data: Dia of sewer: 225 mm; V= 0.75 m/s n=0.013 q= 300 lpcd
Required: Q; S; additional capacity if S is double
Solution : Maximum Capacity: Q = A.VQ = /4 x (225/1000)2 x 0.75 => 0.0298 l/s => 2574.72 m3 /day
Slope using Manning’s Equation V=1/n R 2/3
S½
=> 0.75=1/0.013 (0.225/4)2/3 S ½ S= 0.0044mm-1 No. of Persons
Q = 2574.72 m3 /day = Design flow; Avg flow = 300 lpcdDischarge = lpcd x persons = 2574.72x 1000 = 300 x persons ;
Therefore No. of persons = 8582B. Slope is doubled:
S = 2 x 0.0044= 0.0088
V = 1/0.013 (0.225/4)2/3 (0.0088)1/2 => V = 1.059 m/sQ = AV = /4 x (0.225)2 x 1.059Q = 0.04212 m3 /s = 3639 m3 /dayPersons = 3639 x 1000 / 300Persons = 12130 persons Additional persons =12130-8582 = 3544